Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int g[101][101] = {};
int remain = 0;
int row[101] = {};
int col[101] = {};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) row[i] = -1;
  for (int i = 1; i <= m; i++) col[i] = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> g[i][j];
      if (row[i] == -1)
        row[i] = g[i][j];
      else
        row[i] = min(row[i], g[i][j]);
      if (col[j] == -1)
        col[j] = g[i][j];
      else
        col[j] = min(col[j], g[i][j]);
      remain += g[i][j];
    }
  }
  vector<tuple<bool, int, int>> cmd;
  int cnt{};
  while (remain > 0) {
    int old_remain = remain;
    int maxR = 1;
    for (int i = 2; i <= n; i++) {
      if (row[i] > row[maxR]) maxR = i;
    }
    int maxC = 1;
    for (int i = 2; i <= m; i++) {
      if (col[i] > col[maxC]) maxC = i;
    }
    if ((m > n && row[maxR] != 0) || (col[maxC] == 0)) {
      cmd.push_back({true, maxR, row[maxR]});
      cnt += row[maxR];
      remain -= row[maxR] * m;
      for (int i = 1; i <= m; i++) {
        g[maxR][i] -= row[maxR];
        col[i] = min(col[i], g[maxR][i]);
      }
      row[maxR] = 0;
    } else {
      cmd.push_back({false, maxC, col[maxC]});
      cnt += col[maxC];
      remain -= col[maxC] * n;
      for (int i = 1; i <= n; i++) {
        g[i][maxC] -= col[maxC];
        row[i] = min(row[i], g[i][maxC]);
      }
      col[maxC] = 0;
    }
    if (old_remain == remain) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < cmd.size(); i++) {
    while (get<2>(cmd[i])--)
      cout << (get<0>(cmd[i]) ? "row" : "col") << " " << get<1>(cmd[i]) << endl;
  }
  return 0;
}